Electromotive Force (EMF)

The reading on a high resistance voltmeter, when a cell is connected across it, is $2.0V$. When the terminals of the cell are also connected to a resistance of  $3 \Omega$  as shown in the circuit, the voltmeter reading drops to $1.5V$. The internal resistance of the cell is :

A current of $2 \mathrm{A}$ flows through a $2 \mathrm{\Omega }$ resistor when connected across a battery. The same battery supplied a current of $0.5 \mathrm{A}$ when connected across a $9 \mathrm{\Omega }$ resistor. The internal resistance of the battery is

Two sources of equal emfs are connected in series. This combination is connected to an external resistance $R$. The internal resistances of the two sources are $r_1$ and $r_2\left(r_1>r_2\right)$. If the potential difference across the source of internal resistance $r_1$ is zero then the value of $R$ will be

Two identical cells each of emf $1.5 \mathrm{V}$ are connected in parallel across a parallel combination of two resistors each of resistance $20 \mathrm{\Omega }$. A voltmeter connected in the circuit measures $1.2 \mathrm{V}$. The internal resistance of each cell is :

A cell of $\mathrm{e}\mathrm{m}\mathrm{f}\mathrm{ }E$ and internal resistance $r$ is connected in series with an external resistance $nr$ . Then, the ratio of the terminal potential difference to $\mathrm{e}\mathrm{m}\mathrm{f}$ is $\frac{n}{n+\alpha }$. Write the value of $\alpha$.

The terminal potential difference of a cell when short-circuited is (E=EMF of the cell)

The terminal voltage of a cell in open circuit condition _____is its emf.

A potential difference across the terminals of a battery is $50 \mathrm{V}$ when $11 \mathrm{A}$ current is drawn and $60 \mathrm{V}$ when $1 \mathrm{A}$ current is drawn. The emf and the internal resistance of the battery are

A cell of $\mathrm{e}\mathrm{m}\mathrm{f}\mathrm{ }\mathrm{E}$ and internal resistance $\mathrm{r}$ is connected in series with an external resistance $\mathrm{n}\mathrm{r}$ . Then, the ratio of the terminal potential difference to $\mathrm{e}\mathrm{m}\mathrm{f}$ is -

The potential difference $\left(V\right)$ across the terminals of a battery which is getting charged is always:

The terminal potential difference $\left(V\right)$ of a battery of emf $\left(E\right)$ and internal resistance $\left(r\right)$ which is getting charged with electric current $\left(I\right)$ flowing through it is:

A current $2 \mathrm{A}$ flows through a $2\mathrm{\Omega }$ resistor when connected across a battery. The same battery supplied a current $0.5 \mathrm{A}$ when connected across a $9\mathrm{\Omega }$ resistor. The internal resistance of the battery is

A cell of emf $\epsilon$ and internal resistance $\left(r\right)$ is connected in series with an external resistance $\left(nr\right)$ then the ratio of the terminal potential difference to emf is:

The internal resistance of a cell is the resistance of

A storage battery of emf $8.0 \mathrm{V}$ with internal resistance $0.5\mathrm{\Omega }$ is being charged by a $120 \mathrm{V} \mathrm{D}\mathrm{C}$ supply using a series resistor of $15.5\mathrm{\Omega }.$ What is the terminal voltage of the battery during charging?

First order derivation of thermos emf produced in thermos - couple with respect to temperature gives

Two cells, having the same emf which are connected in series through an external resistance $R$. Cells have internal resistances ${\mathrm{r}}_1$ and ${\mathrm{r}}_2\left({\mathrm{r}}_1>{\mathrm{r}}_2\right)$ respectively. When the circuit is closed, then the potential difference across the first cell is zero. The value of $R$ is:

Assertion: A potential difference measured across cell with a voltmeter is always less than emf of cell.
Reason: Potentiometer is used for emf measurement.